Two fixed charges q_1 =-4.0 mu C (sitting at origin) and q_2 = -6.0 mu C (locate

Question

Two fixed charges q_1 =-4.0 mu C (sitting at origin) and q_2 = -6.0 mu C (located at X = 5.0m). Where on the x-axis a third charge q_3 should be placed so that net force on q_3 is zero? E_net [9/a_2 - a_2/(s - a)^2] = 0 -4/a^2 + -4/(s-a)^2 = 0 4(S-a)^2 = 6a^2 4(S-a)(S-a) = 6a^2 4(2s + a^2 - 10a) = 6d^2 (100 + 4a^2 - 40d) = 6a^2 particle with charge of+4.0 Times 10^-4 C, and mass of 2.0Times10^-9 kg moves at 3.0 he -x direction. It enters a uniform magnetic field of 0.20 T that's points in n. (a) What will be magnitude and direction of the force on this charged magnetic field? Suppose magnetic field exists over a large area, large area charged particle and continues to gyrate in the same circle, (b) Calculate that circle

Explanation / Answer

here

|q1| < |q2|,

so, the third must placed close to q1 and it should be in between q1 and q2.

let x is the distance from q1 where q3 must be placed.

Apply, k*q1*q3/x^2 = k*q2*q3/(5-x)^2

q1/x^2 = q2/(5 - x)^2

(5-x)^2/x^2 = q2/q1

(5-x)/x = sqrt(q2/q1)

(5-x)/x = sqrt(6/4)

5-x = 1.2247*x

5 = 2.2247*x

x = 5/2.22247

= 2.25 m <<<<<-------------Answer

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